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You awaken on a deep space observing station. You do not know its acceleration history but right now there is no gravity and you are billions of light years from the nearest other molecules. A light-year-long series of high speed, high resolution video cameras stretches out left to right as viewed from your observation deck. Each camera has a big LED clock attached to it, which is kept in sync with the clock in your observation post, and which appears in the foreground of every picture. Also there is a yardstick next to each camera so passing observers can measure relative speed.

Your friend awakens aboard a 40m-long spaceship that’s adrift in deep space. He has no memory. After awhile his ship happens to drift in a line exactly parallel to your cameras. His ship has a big LED clock on the side that your cameras can see. His ship itself also has a high-speed camera aboard that records pictures of your cameras’ clocks each time it passes one, and in the foreground of its pictures, the clock attached to his ship is also visible.

Neither of you can ever remember having experienced gravity or acceleration, so it is unknown which of you is the one who accelerated to achieve the speed difference. It’s entirely possible you both accelerated the same amount in opposite directions. Who knows.

From your observation deck, after 1.001001001001001 years, your friend’s ship finally passes the last camera. Upon observing the complete set of photos, you see that your clock and his ship’s clock were exactly in sync in the first photo taken when he initially passed your first camera. Due to the elapsed time, you calculate his velocity was 0.999x the speed of light. Due to time dilation, the clock on the exterior of his ship has only elapsed 16 days, 8 hours, 3 minutes from the first picture you took. Also, his ship only appears to be 1.7884 m in length—4.4471% of the original length.

Meanwhile, your friend’s ship’s clock has also taken photos of your cameras’ clocks as they whizzed past. Finally, there was a quick wifi transmission between his clock and your final clock, over which connection, the final photographs were exchanged.

The Questions

To him, how far would your line of clocks appear to stretch? I calculate 0.044515 LY, due to your length contracting relative to him.

Since you are moving 0.999 c relative to him, I calculate that he would only experience 16 days, 8 hours, 3 minutes having passed while your line of cameras zipped past him. Is this correct? If so, why does he experience a different amount of elapsed time compared to you?

If I’m right then the photo of his own ship that he receives from your final camera will show 16 days, 8 hours, 3 minutes as having elapsed on the outside clock of his ship while your camera’s own foreground clock will indicate 1.001001... years as having passed. Is that what would happen, no matter who was the one that accelerated? Or does some acceleration that occurred in the past affect things? If so, why?

Lastly, the final picture you receive from him would show the same elapsed times as the picture he received from you: his ship has 16 days, 8 hours, 3 minutes, and your final clock has 1.001... years in both pictures. Correct?

If I’m correct on the above points, then I am confused, because it seems asymmetrical.

I would have expected that if you saw his clock elapsed by 16 days, 8 hours, 3 minutes while you experienced 1.001... years, then likewise, he would see:

  • your final clock shows 16 days, 8 hours, 3 minutes passed for you
  • his clock shows a year had passed for him
  • his own calculation of your velocity, using the shrunken yardsticks next to your cameras as a guide, would be 0.999c

However in this scenario, would either of you receive the wifi data exchange? Your final clock would send it after 1.001... years, but to him, your final clock only shows 16 days 8 hours 3 minutes have passed—so how can he be receiving a file that your clock only sends after 1.001... years?

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    $\begingroup$ If you draw the spacetime diagram, this will all be instantly clear. If you don't, it probably never will. $\endgroup$
    – WillO
    May 2, 2019 at 22:46
  • $\begingroup$ @WillO Not everyone finds spacetime diagrams as clarifying as you seem to, especially starting out. My experience is that people starting out still need to be "hand held" through them (often more than once). For most people the "barrier" of their everyday experience of the world is really hard to get past. $\endgroup$ May 3, 2019 at 4:07
  • $\begingroup$ Then they are on their own as far as my time is concerned. This is a good way to distinguish between the interested and the dilettante! $\endgroup$
    – m4r35n357
    May 3, 2019 at 9:45

2 Answers 2

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I am confused, because it seems asymmetrical.

It is asymmetrical because the vehicles are asymmetrical. The length of the 40m craft really has no bearing on the time of the interaction. So for you, the length contraction doesn't matter.

The length of the light-year-long observatory does matter. So the length contraction observed by your friend matters a lot.

Each camera has a big LED clock attached to it, which is kept in sync with the clock in your observation post

Let's amend this sentence:

Each camera has a big LED clock attached to it, which is kept in sync in your frame of reference with the clock in your observation post.

With that in mind, some of your problems should disappear.

Your final clock would send it after 1.001... years, but to him, your final clock only shows 16 days 8 hours 3 minutes have passed

No, your final clock shows a timestamp of 1.001... years, but that does not represent an elapsed time for him. To him, your last clock is not synced to the others and is set almost one year ahead of the first clock.


Addressing later comment:

I was already assuming your light-year-long row of cameras were in your reference frame.

I'm not sure what you mean by this. They're in both reference frames. The clocks only show the same time in the observatory's rest frame though. They show different times in other frames.

What I don’t understand is why both his and your final pictures would show your final clock showing 1.001 years, but his final clock at only ~16 days.

Because they aren't synchronized in the friend's reference frame. They aren't set to the same zero point. Instead of showing the same time, he would describe the clocks as differing by about a year.

In that case, for you, his clock is running slow, and for him, your clock is running fast. But each of you is stationary from your own perspective, so shouldn’t you both perceive the others’ clock to be running slower than your own? If not, then why?

"running slow" can be interpreted in different ways. The clock tick rate and how it's set.

Both of them would observe the other clocks to have a run rate that is slower than their local clocks. What the friend sees is that the end clocks on the observatory are both ticking at a slower rate than his clocks. But the last clock is set to a time about a year in advance of the first clock.

So the friend sees the last clock at the observatory to read "1.001001...years", but interprets this that it's set nearly a year ahead of the first clock. Not that it took a year to get from one side to the other.

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  • $\begingroup$ I was already assuming your light-year-long row of cameras were in your reference frame. What I don’t understand is why both his and your final pictures would show your final clock showing 1.001 years, but his final clock at only ~16 days. In that case, for you, his clock is running slow, and for him, your clock is running fast. But each of you is stationary from your own perspective, so shouldn’t you both perceive the others’ clock to be running slower than your own? If not, then why? $\endgroup$
    – CommaToast
    May 2, 2019 at 22:39
  • $\begingroup$ Responded in the answer. $\endgroup$
    – BowlOfRed
    May 2, 2019 at 22:51
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Length contraction and time dilation are indeed symmetrical effects in special relativity. However, their symmetry might not be apparent if you consider an asymmetrical scenario, as you have.

Time dilation is an effect that arises when you compare the reading on one moving clock with the readings of two separate stationary clocks that it passes. To be doubly clear here, it involves the comparison of elapsed time at a single place in Frame A with the time at two different places in Frame B. To see the symmetry, you must reverse the arrangement and compare the time at a single place in Frame B with the time at two different places in Frame A.

In your thought experiment you consider the time at a single place in the frame of the spaceship with the time at separate places in the frame of the space station. To see the symmetry, you would have to have a light-year-long row of clocks trailing along behind the spaceship; you could then compare the clock on the space station with the clocks towed by the spaceship, and if you were to do that you would find that the time dilation effects were totally symmetrical.

Time dilation arises as a consequence of the relativity of simultaneity. Two frames of reference that are moving relative to each other do not share a common time axis- instead their time axes are tilted relative to each other. That means a plane of constant time in one frame will be a sloping slice through time in the other (and vice versa), the slope being upwards in the direction of motion.

If your space station and spaceship were moving in opposite directions, each towing a long line of cameras and clocks, then if all the cameras towed by the space station were at one moment (in their frame) to take photos of the nearest passing clock in the line towed by the spaceship, they would see that each clock down the spaceship's line was further and further out of synch with the clock on the spaceship. Likewise, if all the cameras towed by the spaceship took a photo at the same instant in their frame, they would see that all the clocks towed by the space station were out of synch.

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